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Setler79 [48]
3 years ago
13

The length of a rectangular deck is 5 times its width. If the deck’s perimeter is 24 feet, what is the deck’s area?

Mathematics
1 answer:
spayn [35]3 years ago
5 0

Answer:

20ft²

Step-by-step explanation:

Let the length = L

Let the width = W

Perimeter of a rectangle = 2L + 2W

Translating the word problem into an algebraic equation, we have;

L = 5W .......equation 1

2L + 2W = 24 ........equation 2

To find the width

Substituting equation 1 into equation 2, we have;

2(5W) + 2W = 24

10W + 2W = 24

12W = 24

W = 24/12

Width, W = 2 ft

To find the length;

Substituting the value of "W" into equation 1, we have;

L = 5W

L = 5*2

L = 10 ft

To find the area of the rectangle;

Area = LW

Area = 10*2

<em>Area = 20 ft²</em>

Therefore, the deck's area is 20 feet square.

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Given the function f(x) = (x-6)^2 + 1

\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

=x^2-12x+37\mathrm{The\: parabola\: params\: are\colon}a=1,\: b=-12,\: c=37x_v=-\frac{b}{2a}x_v=-\frac{\left(-12\right)}{2\cdot\:1}\mathrm{Simplify}x_v=6y_v=6^2-12\cdot\: 6+37

Simplify:

y_v=1

\mathrm{Therefore\: the\: parabola\: vertex\: is}\mleft(6,\: 1\mright)\mathrm{If}\: a\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}a=1\mathrm{Minimum}\mleft(6,\: 1\mright)\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

y=x^2-12x+37\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}a=1,\: b=-12x=\frac{-\left(-12\right)}{2\cdot\:1}\mathrm{Refine}

Axis of simmetry : x=6

The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.

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1 year ago
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